Homotopy equivalence between topological mapping cone and the algebraic mapping cone

In the Wikipedia article of mapping cones there’s the following:

This complex is called the cone in analogy to the mapping cone (topology) of a continuous map of topological spaces $$\phi \colon X \rightarrow Y$$: the complex of singular chains of the topological cone $$\mathrm{cone}(\phi)$$ is homotopy equivalent to the cone (in the chain-complex-sense) of the induced map of singular chains of $$X$$ to $$Y$$.

I would like to see an explicit homotopy equivalence. We clearly have a chain map from $$C_{n-1}(X) \oplus C_n(Y)$$ to $$C_n(\mathrm{cone}(\phi))$$ (for an element $$\sigma \in C_{n-1}(X)$$ triangulate the cone $$\sigma * \{1\}$$; for an element in $$C_n(Y)$$ just map it to the base of the mapping cone). I couldn’t think of maps going in the other direction. Could someone help?

• Interesting question. I don't think a map $|\Delta^n|\to C_\phi$ need induce maps $|\Delta^n|\to Y$ or $|\Delta^{n-1}|\to X$ in any direct way, there's probably some 'swindle'. Dec 23, 2022 at 12:04

$$\DeclareMathOperator{\C}{Sing}\DeclareMathOperator{\Cone}{Cone}$$ Added later. Probably the correct thing is to say that $$C_\varphi$$ is a homotopy pushout (the other map being the trivial map from $$X$$ to a point) in spaces while $$\Cone(f)$$ is a homotopy pushout in chain complexes, and that $$\C$$ preserves homotopy pushouts up to homotopy. I thought this was true but reading this post leaves me doubting.